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Re: [Amps] Line Sections As Plate Lines

To: JMLTINC@aol.com
Subject: Re: [Amps] Line Sections As Plate Lines
From: David Kirkby <david.kirkby@onetel.net>
Date: Sun, 06 Mar 2005 04:33:24 +0000
List-post: <mailto:amps@contesting.com>
JMLTINC@aol.com wrote:

>Hi again guys-
>
>Sorry if I was not clear about my question. Let me try it from a different 
>angle. Specifically:
>
>How do you calculate the length for a 1/2wl plate line????
>That is, what is the formula?
>
>I already know:
>Freq = 144Mhz
>Zo = 87 ohms
>Ctotal (tube, strays, tune) = 74 ohms
>
>Thanks-
>John, N9RF
>  
>

I assume you mean Ctotal is 14.8pF, giving a reactance of  74 Ohms.  
Anyway, forget the numbers - a general formula is better. I assume the 
tube is at one end of a line, and the other end is a high impedance too.

The half-wave line is open circuited at one end, with the tube at the 
other. But there are an infinite number of combinations of line length 
that can be made to resonate, in different modes (0.5 lambda, 1 lambda, 
1.5 lambda etc), all open circuit at one end. Hence care is needed when 
getting a formula that you get the right solution. I seem to find a very 
different one from the results you give, but I'll leave it for others to 
find a fault if I have made an error.

I had a go at deriving a formula. I did not derive it from first 
principles, but don't use an 'amateur knudges' that might be valid, but 
might not be.  Reducing my solution to its simplest useful form I got:

2 Pi f Ctotal  Zo ==  - Tan(2 Pi f  L / c)

where L is the length of the line you want to find in m.
c is the velocity of light 3 x 10^8 m/s
f is the frequency in Hz.
Pi=3.14159
Ctotal is the output capacitance in Farads.
Zo is the impedance of the transmission line.

Attempting to take the ArcTan's of both sides never seemed to work to 
well for me, although it would have in principle given a simpler 
expression. But I'm pretty poor at Maths, so it's probably me - either 
that or a bug in Mathematica. I might have a go in Matlab later and see 
if I do any better with that, but I'm not such a competent user of that.

I understand when doing this you loose all except one solution, but I 
seem to get a totally wild solution. Anyway, forgetting about trying to 
invert fully, but instead doing a simple numerical solution (which you 
could do on a calculator with a bit of thought about it), I tried it in 
Mathematica.

Just for the record, 'FindRoot' is a function in Mathematica that looks 
a value of L such that the LHS == RHS. There are an infinite number of 
solutions, so I tell it to look around 0.6m, as I know there is one 
around there, which I *believe* is the right one, but could be wrong.


FindRoot[ 2 Pi 144000000 14.8 10^-12    89 ==  - Tan [2 Pi L  144 10^6/ 
(3 10^8) ], {L, 0.6}]
{L -> 0.752312}

which indicates a solution for .752m. If you insist on working in 
inches, I'll leave you the job of converting it!

There are also solutions at -0.289m (not physically realisable, as 
length <0) and at 1.794m, which is a higher mode. There's nothing wrong 
with having negative solutions - mathematically they are valid, but they 
are a bit hard to make.

Proof of my formula - there may be an error here.
****************************************

The *general* equation for the input impedance of a lossless 
transmission line of length 'L' , characteristic impedance Zo, 
terminated in a load impedance of Zload is:

Zin=Zo * (Zload cos (beta * L ) + j Zo Sin(beta * L))/(Zo Cos(beta * L) 
+ j Zload Sin(beta * L))

where beta = 2*pi/lambda
j is the complex operator.
and c=3 10^8 m/s.

That's valid for real and complex loads.

If Zload =0 (short circuit, which it would be for a quarter wave line), 
then that complex equation reduces to
Zin = j Zo tan (beta L) - this is just stated so you can redo it for a 
quarter wave if you want.

If Zload=Infinity (open circuit, which is what you will do for a 
half-wave line), then that big complex equation reduces to

Zin = - j cot[beta L]  which simplifies things quite a bit.

( Note: cot(x) = 1/tan(x) )

The  reactance of the tube is

Xc = -j/(2 Pi f Ctotal))

But for resonance, Xc = - Zin, since you must balance the capacitive 
(negative) reactance of the tube with an inductive (positive) reactance 
from the transmission line. (The signs of the reactance must be opposite).

Hence -j/(2 Pi f Ctotal) ==  j Zo cot (beta L)  for resonance

imaginary parts cancel, so

-1/(2 Pi f Ctotal) == Zo cot (beta L)

Inverting.

2 Pi f Cotal  Zo ==  - Tan(beta L)


using the fact beta = 2 Pi/lambda and 1/lambda = f/c, gives:

2 Pi f Ctotal  Zo ==  - Tan(2 Pi f  L / c)

There might be an error in that lot, but I'm sure someone will pick it 
to bits (I hope so anyway). It is of course a theoretical computation, 
and ignores things such as the fact the transmission line is short, so 
its not really too valid to give it a characteristic impedance, when end 
effects are significant.

-- 
Dr. David Kirkby, 
G8WRB

Please check out http://www.g8wrb.org/ 
of if you live in Essex http://www.southminster-branch-line.org.uk/



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